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Suppose we have a particle in an infinite potential well, with $V(x) = 0,\space 0< x < a $ and infinity everywhere else.

Now suppose we have a perturbation on the LHS of the well: $V_1(x) = v, \space 0 < x < \frac{a}{2}$.

By first order perturbation theory, all energy levels are shifted by the same amount:

$$\Delta E = \langle E_n|v|E_n\rangle = \frac{v}{2} $$

Ok all is good.

By considering the perturbed wavefunction:

$$ \psi_n = \phi_n + \sum_{n \neq k} \frac{\langle \phi_k|V_1|\phi_ n\rangle}{E_n - E_k} $$

How do I make use of the condition that $v << E_2 - E_1 $ to show the results above is correct?

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  • $\begingroup$ Isn't $\sum_{n \neq k} \frac{\langle \phi_k|V_1|\phi_ n\rangle}{E_n - E_k} = 0$ ? Since states are orthogonal? $\endgroup$ – user44840 Jun 5 '14 at 12:19
  • $\begingroup$ $V_1$ changes $|\phi_n\rangle$, so in general the product it's no longer orthogonal to $|\phi_k\rangle$. $\endgroup$ – Ruslan Jun 5 '14 at 12:22
  • $\begingroup$ $V_1 (x) = v$ is a constant, so I think it can be taken out of the bra and ket $\endgroup$ – user44840 Jun 5 '14 at 12:25
  • $\begingroup$ Dot product doens't make any sense if you only integrate over one part of configuration space. In this case you'd not get any orthogonality. Just try checking if e.g. $\sin 2\pi x$ is orthogonal to $\sin \pi x$ on $x\in[0,1]$. It's not, although on $x\in[0,\pi]$ it is. $\endgroup$ – Ruslan Jun 5 '14 at 12:27
  • $\begingroup$ $V_1$ is not constant, it is $v$ for $x<a/2$ and $0$ otherwise. This means that the integral of $\phi_k$ with $\phi_n$ only goes from $0$ to $a/2$, and orthogonality is not guaranteed. $\endgroup$ – Davidmh Jun 5 '14 at 12:29
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It's actually an interesting problem you raise, so let me try to expand a bit on my comment.

In general, the program for “showing that [first-order] perturbation theory (PT) works” is the following:

  1. Solve the perturbed problem exactly, yielding solutions $H\psi_n = E_n\psi_n$.
  2. Differentiate the results by the perturbation parameter, i.e. compute $dE_n/dv$ and $d\psi_n(x)/dv$.
  3. Compare this to the PT results $E_n^{(1)}$ and $\psi_n^{(1)}(x)$.

In the present case, this will not be possible analytically. Write an Ansatz for the full solution, split in two regions, (L) $V(x)=v$, and (R) $V(x)=0$. As in the regular infinite potential well, for each region you can eliminate one of the linearly independent solutions by the boundary conditions $\psi_L(0) = \psi_R(a) = 0$.

You are left with three parameters (an “amplitude” for each region, and the energy) and three relations to fix them ($\psi_R(a/2) = \psi_L(a/2)$, $\psi'_R(a/2) = \psi'_L(a/2)$, $\int\!dx\,\psi(x) = 1$). Simple enough in principle, but not analytically solvable in this case because the energy is given by an implicit transcendental equation.

The reason I say your problem is interesting is that, as you increase $v$, it will change the character of the lower-lying states. In the unperturbed case, all states are “oscillatory”, albeit cut off at the edges. In the perturbed case, this will still be true on the right side, but once $v$ is big enough, you will get states that decay exponentially on the left. A sensible guess would be that PT breaks down for these states.

As for $V_{nm} = \langle\phi_n|V|\phi_m\rangle$ and orthogonality: The $\phi$ are even/odd functions with respect to the well center. Split the integral $\langle\phi_n|\phi_m\rangle = \int_0^a\!dx\phi_n\phi_m = L + R$ into two parts corresponding to left and right side. As you say, $L + R = 0$; and $V_{nm} = vL$. If $n$ and $m$ have the same parity, $L=R$ and so $V_{nm} = 0$; but if the parity is different, $L=-R$ and $V_{nm}$ is non-zero.

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